Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long n, m, ans = 0, ans1 = 0;
int a[105][105], cop[105][105], row[105], col[105];
vector<int> r, c, r1, c1;
int main() {
  cin >> n >> m;
  for (int i = 0; i < 105; i++) {
    row[i] = col[i] = 1000;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
      cop[i][j] = a[i][j];
      row[i] = min(row[i], a[i][j]);
      col[j] = min(col[i], a[i][j]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      a[i][j] -= row[i];
    }
    for (int t = 0; t < row[i]; t++) r.push_back(i + 1);
    ans += row[i];
  }
  for (int i = 0; i < 105; i++) {
    col[i] = 1000;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      col[j] = min(a[i][j], col[j]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i][j] != col[j]) {
        cout << "-1";
        return 0;
      }
    }
  }
  for (int i = 0; i < m; i++) {
    ans += col[i];
    for (int t = 0; t < col[i]; t++) c.push_back(i + 1);
  }
  for (int i = 0; i < 105; i++) {
    row[i] = col[i] = 1000;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      row[i] = min(row[i], cop[i][j]);
      col[j] = min(col[j], cop[i][j]);
    }
  }
  for (int j = 0; j < m; j++) {
    for (int i = 0; i < n; i++) {
      cop[i][j] -= col[j];
    }
    for (int t = 0; t < col[j]; t++) c1.push_back(j + 1);
    ans1 += col[j];
  }
  for (int i = 0; i < 105; i++) {
    row[i] = 1000;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      row[i] = min(cop[i][j], row[i]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (cop[i][j] != row[i]) {
        cout << "-1";
        return 0;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    ans1 += row[i];
    for (int t = 0; t < row[i]; t++) r1.push_back(i + 1);
  }
  if (ans <= ans1) {
    cout << ans << endl;
    for (int i = 0; i < r.size(); i++)
      cout << "row"
           << " " << r[i] << endl;
    for (int i = 0; i < c.size(); i++)
      cout << "col"
           << " " << c[i] << endl;
  } else {
    cout << ans1 << endl;
    for (int i = 0; i < r1.size(); i++)
      cout << "row"
           << " " << r1[i] << endl;
    for (int i = 0; i < c1.size(); i++)
      cout << "col"
           << " " << c1[i] << endl;
  }
}